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University of Waterloo
Faculty of Mathematics
Centre for Education in
Mathematics and Computing
Senior Math Circles
November 25, 2009
Graph Theory II
Planar Graphs
A graph that can be drawn in R2 without any edges crossing
is planar. An example of a graph which is not planar is the
complete graph on 5 vertices (the graph
in which every vertex is
adjacent to every other vertex; all 52 = 10 edges are present).
crossing
F4
F5
F1
F3
F6
F2
When we draw a planar graph, it divides R2 into different
regions which are called faces. When talking about the faces of
a graph, the infinite outer face is included. For example, the
diagram here of a cube graph has 6 faces, denoted F1 , . . . , F6
with F6 the infinite outer face.
4-colour theorem: The faces of any planar graph can be coloured using at most 4 colours in total
such that any two adjacent faces are not the same colour.
– The theorem was proposed in 1852, and was popularized by Cayley.
– It was proven in 1879 & 1880.
– Flaws were found in those proofs in 1890 & 1891.
– It remained open again until it was finally proven in 1976 by Appel & Haken.
−→ computers were involved in checking a finite number of cases of proof.
−→ proof overview: give a list of reducible configurations, show that induction can be used
whenever a reducible configuration is found, and show that every graph contains a reducible
configuration.
−→ people are still coming up with simpler and more transparent proofs today.
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Note that we only disallow faces which share an edge to be coloured the
same — it is okay if like-coloured faces share a vertex. Otherwise we can
come up with graphs (like the one pictured) which take an arbitrarily
large number of colours.
In the problems at the end of this handout, we show that a weaker “6-colour theorem” is reasonably
easy to prove.
In a connected planar graph, Euler’s formula states that
the number of vertices − the number of edges + the number of faces = 2
where the number of faces includes the infinite outer face. We will often abbreviate this as
|V | + |F | − |E| = 2
where |V | is the number of vertices, |F | is the number of faces, and |E| is the number of edges.
Dual Graphs.
Every planar graph has a dual graph. For any connected planar graph G, the dual of its dual is again
G. To obtain the dual of a graph (e.g., the cube graph shown with black vertices below), we put a
white vertex in every face of G. Then, whenever two faces of G are next to each other, we connect
the corresponding two white vertices with a dual edge. Note that we can delete the original graph
and just deal with the dual graph by itself, as shown in the figure below.
Planar Dual
−→
−→
Original Graph
We always have the following:
number of original faces = number of dual vertices
number of original edges = number of dual edges
number of original vertices = number of dual faces
For example, to prove the first fact, just recall that we put one dual vertex in each face of the original
graph. Another fact is that, if v is a vertex of the original graph of degree d, then the dual face
corresponding to v has exactly d sides (see the diagram for an idea of the proof).
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In the example above, the original graph had degree 3 at every vertex and every face was a square;
this implies that the dual graph has triangular faces and degree 4 at every vertex, i.e. the dual graph
is a octahedron.
Exercise. Compute the dual graphs of the tetrahedron, icosahedron, and dodecahedron. How are
these graphs related to each other?
Duality has a nice effect on the 4-colour theorem. Instead of colouring faces of a planar graph, we
can think about colouring vertices of the dual graph in such a way that adjacent vertices get different
colours. (Why is this nice? For example, we can’t colour the faces of non-planar graphs since they
don’t exist, but we can colour their vertices.) We call an assignment of colours to vertices a proper
colouring if every adjacent pair gets different colours.
Exercise. What is the least total number of colours required to properly colour the following graph?
The graph in the exercise above is usually denoted K3,3 and called the complete bipartite graph, since
it has two (“bi-”) groups (“parts”) of three vertices ( 3,3 ) such that each vertex is connected to all
vertices in the other part (“complete.”)
Breadth-First Search. Here is a natural question, related to applications like Google Maps. I give
you a graph containing vertices A and B. What is the least number of steps (along edges) that will
get me from vertex A to vertex B?
In the example below, we see that there is a path P (bold, blue edges) of length 4. But how can we
actually prove this is the shortest path?
A
B
We now give the most common all-purpose idea to find shortest paths in graphs, which is called
breadth-first search. The idea is that we will label every vertex according to its distance from A, i.e.
the label of v is the minimum number of steps to get from A to v. We need to compute all the small
numbers first before computing the large numbers, and the term breadth-first refers to the fact that
we are sort of searching outwards from A in layers.
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First, we label A with the number 0. Then we label all neighbours of A with 1. Next, for all vertices
v with label 1, for every unlabeled vertex w adjacent to v, we label w with 2. Then the unlabeled
neighbours of 2-labeled vertices become 3-labeled, et cetera. You can prove by induction on the
distance from A that this assigns the correct distance label to every vertex. The diagram below gives
an example.
2
2
1
1
2
2
2
2
−→
A 0
3
A 0
2
3
2
1
1
2
2
3
3
3
B
B
2
1
2
2
−→
3
A 0
3
2
3
1
2
3
4
3
4
4
B
Implicit Search & Meet in the Middle. Graphs are hidden everywhere. In some cases, a “graph
search” problem is the most natural way to think about a problem which says nothing about graphs!
An example is a word ladder puzzle. There are lots of variations on this idea, so we describe it in
full detail. You are given a “start word” and an “end word,” both 4 letters long. (E.g. COLD for
the start and HEAT for the end.) The goal is to quickly transform the start word into the end word,
subject to the following rules: (i) you can only change one letter at a time, and (ii) all intermediate
words must be actual English words. (So you can’t change COLD into ZOLD.)
In some cases, word ladders are easy. The given example has a solution with 4 transformations:
COLD-HOLD-HELD-HEAD-HEAT. This is clearly optimal since we need to change each letter once.
But for more complex situations, this reasoning won’t always figure out the shortest solution.
Here is a way of thinking about the problem in terms of graphs: define a vertex for every 4-letter
English word, and connect two words by an edge whenever they differ only in one letter. Then we
are just looking for a shortest path!
Exercise. Use breadth-first search to find a shortest path from POST to POTS.
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Another conceivable situation is that you have a Rubik’s cube that you want to solve using the least
number of possible operations (turns of a side). You could model this by a graph with a vertex for
each possible state/picture of the cube, and vertices linked by all possible face turns. In principle,
if you know how to do some programming, and a little graph theory, this will allow you to find the
shortest path between the original (mixed) state of the cube and the final (solved) state where all
faces have one colour.
We say “in principle” above because there are a lot of states to check, which takes up a lot of memory
and uses a lot of processor time, even for today’s modern computers. This complexity is important:
if you ask Google Maps how to get to the supermarket because you are hungry, if it takes 2 weeks to
compute the answer, then you may have starved by then. Therefore, it is useful to have algorithms
which have better performance than breadth-first search. A simple improvement to breadth firstsearch is the meet in the middle algorithm which interleaves a forwards breadth-first search from
the start with a backwards breadth-first search from the end. We start whenever both searches have
a common member.
For example, let’s find a shortest path from FOAM to FOOD. We start by searching one layer around
FOAM: it is adjacent to ROAM, FORM, and FOAL. Then we go to FOOD and search one layer
around it: it is adjacent to MOOD, FORD, FOOL, FOOT. Remark: as we go along, we are drawing
tree structures which grow outwards from FOAM and FOOD, which are usually called breadth-first
search trees.
MOOD
FOAL
FORD
FOOD
ROAM
FOAM
FOOL
FORM
FOOT
We just grew the FOOD tree so we go back to add a layer to the FOAM tree, which is pictured
below. At this point the word FORD is in both trees so we are done: a shortest path is FOAMFORM-FORD-FOOD.
REAM
ROOM
ROAR
ROAD
ROAM
NORM
FARM
FOAM
FORM
FIRM
FORD
FOAL
etc.
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In practice, this meet in the middle search can result in a huge savings. Imagine some search problem
where the distance was 16. Suppose for the sake of estimating that the tree increases in size by a
factor of 3 nodes at each level. Then a BFS would result in a tree of size about 316 ≈ 43 million, but
meet-in the middle grows two trees each of size about 38 ≈ 6500, which is a huge savings in work.
Proofs of Euler’s Formula.
To conclude, since Euler’s formula is one of the most well-used theorems about planar graphs, we
should do it justice by actually figuring out how to prove it. We sketch two proof ideas here.
Proof 1. Use induction. It is not so easy to figure out an appropriate induction variable, but a
natural base case is trees: any tree has just one face, so |V | + |F | − |E| = |V | + 1 − (|V | − 1) = 2 as
needed. Then complete the proof by arguing that whenever we add an edge to the graph, it increases
the number of faces by exactly 1.
Proof 2. Use Problem 1 in the exercises below for a direct proof.
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Problems
1. Consider a connected planar graph G. Let T be a spanning tree of G. Let X be the set of all
edges not in T .
(a) Using the fact that T connects all vertices, show that in the dual graph G∗ , X is acyclic.
(b) Using the fact that T is acyclic, show that in the dual graph G∗ , X connects all vertices.
(Remember vertices of G∗ correspond to faces of G.)
(c) Deduce, using (a) and (b), that X is the edge set of a spanning tree of G∗ .
2. A colouring of the vertices of G is proper if, for all edges uv of G, the colour of u and the colour
of v are different. The 4-colour theorem states that “every planar graph has a proper colouring
using at most four colours in total.” We will prove the weaker but easier statement that “every
planar graph has a proper colouring using at most six colours in total.”
Part 1. Assume the following statement is true:
Every planar graph has at least one vertex v such that degree(v) ≤ 5.
(?)
Assuming that (?) is true, show by induction that every planar graph has a proper colouring
using at most six colours in total. (Hint: deleting v from G leaves a planar graph.)
Part 2. We will prove (?).
(a) If G has no cycles, show that G has a vertex of degree at most 5. (Hint: 1 ≤ 5.)
(b) If G has any cycles, then it is easy to argue that every face has at least three edges.
Let |E| denote the number of edges of G and |F | denote the number of faces of G. Show that
|F | ≤ 23 |E|.
(c) Suppose for the sake of contradiction that (?) is false. Then there is a counterexample
planar graph G in which every vertex has degree at least 6. Show that in this graph, |V | ≤ 13 |E|,
where |V | denotes the number of vertices of G.
(d) Complete the proof of (?) by using Euler’s formula.
3. Use breadth-first search to prove one direction of last week’s problem #6: if a graph has no
cycles of odd length, construct a proper colouring using at most two colours.
v
4. Find the furthest vertex (or vertices) from v in the
graph on the left below. (It is called the Nauru
graph.)
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s
t
5. Modify the breadth-first search and meet-inthe-middle search algorithms to work on directed graphs. As an example, find the shortest distance from s to t in the directed graph
on the right above.
6. The 4-letter word graph has vertex set equal to the set of all 4-letter English words, and edges
connecting two words when the letters in three positions are the same. (For example, MATH
and MOTH are adjacent, POST and POTS are not adjacent, and an example of a path is
COLD-HOLD-HELD-HEAD-HEAT.)
(a) Find a cycle of length 3.
(b) Find a shortest path between JUNK and GOLD.
(c) Prove the graph is not planar (hint: use the 4-colour theorem and consider all words of
the form ?ATS).
Further resources:
• The “draw this figure without tracing any line more than once” puzzle from the first lecture
is basically a question about whether the associated graph has a path which visits every edge
once (but maybe some vertices more than once). It is called an Eulerian path. An Eulerian
path exists if and only if the graph is connected, and at most two vertices have odd degree.
An Eulerian cycle visits every edge exactly once and ends where it starts; an Eulerian cycle
exists if and only if the graph is connected, and all vertices have even degree. See a proof at
http://www.stanford.edu/class/cme305/Lectures/Lecture1.pdf